Question

Find an equation of the plane that passes through the point and contains the given line.
(3, 4, 5)
x = 5t, y = 3 + t, z = 4 - t

Answer 1

the given line gives you one vector on the plane <5,1,-1>

Answer 2

a(x-3)+b(y-4)+c(z-5)

<5,1,-1><a,b,c>=0

Answer 3

it also gives you a point on the plane (0,3,4)

Answer 4

use that point and the point given to create another vector with and cross the vectors to get a "normal" vector to the plane

Answer 5

gonna have to check my work since its been awhile since I tried crossing vectors :)

(0,3,4) moves to
(3,4,5)
------
<3,1,1> is the vector between the given points
<5,1,-1> is the vector from the line equations

< i, j ,k>
<3, 1, 1>
<5,1,-1>

(1.-1 - 1.1)i + (1.5 - 3.-1)j + (3.1 - 5.1)k

Answer 6

<-2,8,-2> if i did it right; lets check by dotting it to the other 2

<-2,8,-2>
< 3, 1, 1>
----------
-6 +8 -2 = 0 mighta got lucky



<-2,8,-2>
< 5, 1,-1>
----------
-10+8+2 = 0 .... looks good

Answer 7

cant recall if we "simplify" the vector: -2<1,-4,1> maybe?

1,-4,1 1,-4, 1
3, 1, 1 5,1,-1
------ -------
3-4+1 = 0 5-4-1 = 0 still works :)

Answer 8

so lets take our point and apply the normal vector to it to define the plane:

1(x-3) -4(y-4) +1(z-5) = 0
or maybe
x -4y +z +8 = 0

Answer 9

we should get the same results for the second equation regardless of the point we use ... lets try using the point given in the line stuff (0,3,4)

1(x-0)-4(y-3)+1(z-4) = 0

x +4y +z (+12 -4 ) = 0

x +4y +z +8 = 0 ... appears to be good

Answer 10

Thanks man

Answer 11

youre welcome :) I see a typo in that last bit .. should be -4y, but its good ;)